The ellipse x^2+4y^2=4 is inscribed in a rectangle aligned with the coordinate axes, which…

Question

The ellipse $x^2+4y^2=4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point $(4,0)$. Then the equation of the ellipse is:

Accepted Answer

$x^2+4y^2=4\Rightarrow\dfrac{x^2}{4}+\dfrac{y^2}{1}=1$, so its bounding rectangle (touching the ellipse at its axis endpoints) has vertices $(\pm2,\pm1)$.

Let the outer ellipse be $\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1$, passing through the rectangle's vertex $(2,1)$ and through $(4,0)$:
$$\frac{16}{A^2}=1\Rightarrow A^2=16,\qquad \frac{4}{16}+\frac1{B^2}=1\Rightarrow B^2=\frac43.$$
Equation: $\dfrac{x^2}{16}+\dfrac{y^2}{4/3}=1\Rightarrow x^2+12y^2=16.$

*(The source's recorded answer letter did not match this derivation; option A, $x^2+12y^2=16$, is the value consistent with the computation above.)*

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